Answer
a. 2NaOH + $H_{2}CO_{3}$ $\longrightarrow$ $Na_{2}$C$O_{3}$+2$H_{2}$O
b. Si$Cl_{4}$ + 2$H_{2}$O $\longrightarrow$ Si$O_{2}$+4HCl
c. $Ca_{3}(PO_{4})_{2}$ +8C $\longrightarrow$ $Ca_{3}$$P_{2}$ + 8CO
d. 2$H_{2}$S + 3$O_{2}$ $\longrightarrow$ 2S$O_{2}$ + 2$H_{2}$O
e. 2$N_{2}O_{5}$ $\longrightarrow$ 4N$O_{2}$ + $O_{2}$
Work Step by Step
The strategy requires to find the coefficient to balance the chemical equation. First look for atoms of elements that occur in only one substance on each side of equation.
Begin by balancing these atoms to get to the balanced chemical equation.
Count the atoms of each element in each side of equation, to see if they are in the same number in each side.
a. NaOH + $H_{2}CO_{3}$ $\longrightarrow$ $Na_{2}$C$O_{3}$+$H_{2}$O
Balance Na atoms by adding 2 before NaOH
2NaOH + $H_{2}CO_{3}$ $\longrightarrow$ $Na_{2}$C$O_{3}$+$H_{2}$O
Balance H atoms as there are 4 H atoms on the left and 2 on the right. Add 2 before $H_{2}$O
2NaOH + $H_{2}CO_{3}$ $\longrightarrow$ $Na_{2}$C$O_{3}$+2$H_{2}$O
Count the atoms of each elements and find our that we have balanced the equation.
b. Si$Cl_{4}$ + $H_{2}$O $\longrightarrow$ Si$O_{2}$+HCl
Balance Cl atoms by adding 4 before HCl
Si$Cl_{4}$ + $H_{2}$O $\longrightarrow$ Si$O_{2}$+4HCl
Balance O atoms as there is 1 O atom on the left and 2 on the right. Add 2 before $H_{2}$O
Si$Cl_{4}$ + 2$H_{2}$O $\longrightarrow$ Si$O_{2}$+4HCl
Count the atoms of each elements and find our that we have balanced the equation.
c. $Ca_{3}(PO_{4})_{2}$ +C $\longrightarrow$ $Ca_{3}$$P_{2}$ + CO
we have to balance O atoms as there are 8 ( 4x2) on the left and 1 on the right. Add 8 before CO:
$Ca_{3}(PO_{4})_{2}$ +C $\longrightarrow$ $Ca_{3}$$P_{2}$ + 8CO
Balance C atoms as there is only 1 on the left, and there are 8 on the right. Add 8 before C:
$Ca_{3}(PO_{4})_{2}$ +8C $\longrightarrow$ $Ca_{3}$$P_{2}$ + 8CO
Count the atoms of each elements and find our that we have balanced the equation.
d. $H_{2}$S + $O_{2}$ $\longrightarrow$ S$O_{2}$ + $H_{2}$O
Balance O atoms as there are 2 on the left and 3 on the right. To balance add 3 before $O_{2}$ , 2 before S$O_{2}$ and 2 before $H_{2}$O:
$H_{2}$S + 3$O_{2}$ $\longrightarrow$ 2S$O_{2}$ + 2$H_{2}$O
Now we have 6 ( 3x2) O atoms on the left and 6 ( 2 x2 + 2) O atoms in the right.
Balance S atoms as there is 1 on the left and 2 on the right. Add 2 before $H_{2}$S
2$H_{2}$S + 3$O_{2}$ $\longrightarrow$ 2S$O_{2}$ + 2$H_{2}$O
Count the atoms of each elements and find our that we have balanced the equation.
e. $N_{2}O_{5}$ $\longrightarrow$ N$O_{2}$ + $O_{2}$
Balance N atoms as there are 2 on the leftand 1 n the right. Add 2 before N$O_{2}$
$N_{2}O_{5}$ $\longrightarrow$ 2N$O_{2}$ + $O_{2}$
Balance O atoms as there are 5 on the left and 6 ( 2x2 +2) on thhe right. Add 2 before $N_{2}O_{5}$ and add 4 before N$O_{2}$
2$N_{2}O_{5}$ $\longrightarrow$ 4N$O_{2}$ + $O_{2}$
Count the atoms of each elements and find our that we have balanced the equation.
